# $f(x)$ is a differentiable function on the real line such that $\lim_{x \rightarrow \infty} f(x)=1$ and $\lim_{x \rightarrow \infty} f~'(x)=\alpha$.

$$f(x)$$ is a differentiable function on the real line such that $$\lim_{x \rightarrow \infty} f(x)=1$$ and $$\lim_{x \rightarrow \infty} f~'(x)=\alpha$$. Then :

$$(A) ~\alpha$$ must be $$0$$

$$(B) ~\alpha$$ need not be $$0$$ but $$|\alpha|<1$$

$$(C) ~\alpha > 1$$

$$(D) ~\alpha < -1$$

Attempt:

$$f~'(x) = \lim_{h \rightarrow 0} \dfrac {f(x+h)-f(x)}{h}$$

$$\lim_{x \rightarrow \infty} f~'(x) =\lim_{x \rightarrow \infty} \lim_{h \rightarrow 0} \dfrac {f(x+h)-f(x)}{h}$$

$$\lim_{x \rightarrow \infty} f~'(x) = \lim_{h \rightarrow 0} \lim_{x \rightarrow \infty}\dfrac {f(x+h)-f(x)}{h}\tag{1}$$

$$\lim_{x \rightarrow \infty} f~'(x) = \lim_{h \rightarrow 0}\dfrac {1-1}{h} = 0$$

Hence $$\alpha$$ must be equal to $$0$$.

Is step $$(1)$$ legal?

I mean, can we interchange the two limits?

When can we not interchange the two limits and when can we?

Thank you very much for your help in this regard.

• The conclusion is correct: $\;\alpha =0\;$, yet step (1) needs justification and, in the general case, it can fail big time as you can see in some examples here: math.stackexchange.com/questions/15240/… – Timbuc Apr 29 '15 at 6:59

Justify each of the following steps:

We can try the following: for $\;n\in\Bbb N\;$ :

$$f(n+1)-f(n)=\frac{f(n+1)-f(n)}{(n+1)-n}=f'(c_n)\;\;,\;\;\;c_n\in (n,\,n+1)$$

using the MVT (Lagrange's Theorem) for differentiable functions.

From this it follows that $\;c_n\xrightarrow[n\to\infty]{}\infty\;$, and thus

$$\alpha=\lim_{n\to\infty}f'(c_n)=\lim_{n\to\infty}\left(f(n+1)-f(n)\right)=1-1=0$$

• Got it. Thank you for your answer :) – MathMan Apr 29 '15 at 9:14